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3 years ago

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PLS HELP!I will give brainliest! When producing hydrogen iodide, the energy of the reactants is 581 kJ/mol, and the energy of th

e products is 590 kJ/mol. The equation is shown.
H2 + I2 → 2HI

What is the total energy of the reaction? Is this an endothermic or exothermic reaction?
A. –9 kJ/mol, exothermic
B.9 kJ/mol, exothermic
C. 9 kJ/mol, endothermic
D. –9 kJ/mol, endothermic

1 answer:

Veronika [31]3 years ago

4 0

Answer:

Here, we are required to determine the total energy of the reaction and determine if the reaction is an endothermic or exothermic reaction.

The correct answer is option C.

First, we need to determine the energy of the reaction.

The energy of the reaction is the change in enthalpy between the product and reactants.

Change of Enthalpy,

Hreaction = Hproduct - Hreactant.

Therefore, for the reaction above, the change in enthalpy is:

Hreaction = 590kJ/mol - 581kJ/mol.

Hreaction = 9kJ/mol.

Hence, since the reaction has an enthalpy change of 9kJ/mol, the reaction is endothermic (i.e energy is absorbed).

Explanation:

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Answer:

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Explanation:

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The property of carbon dioxide to get excited by infrared electromagnetic radiation is what qualifies it as a greenhouse gas. When infrared from the earth's surface is reflecting back to space, some of the radiation is absorbed and remitted by the carbon-dioxide molecules in the atmosphere. This causes a phenomenon called greenhouse effect that causes the atmosphere to be relatively warmer. The more the carbon dioxide molecules the more the greenhouse effect.

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Explanation:

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A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.01962 mol of

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Answer:

See explanation.

Explanation:

Hello,

In this case, we can show how the empirical formula is found by following the shown below procedure:

1. Compute the moles of carbon in carbon dioxide as the only source of carbon at the products:

$n_C=0.01962molCO_2*\frac{1molC}{1molCO_2} =0.01962molC$

2. Compute the moles of hydrogen in water as the only source of hydrogen at the products:

$n_H=0.01961molH_2O*\frac{2molH}{1molH_2O}=0.03922molH$

3. Compute the mass of oxygen by subtracting the mass of both carbon and hydrogen from the 0.4647-g sample:

$m_O=0.4647g-0.01962molC*\frac{12gC}{1molC}-0.03922molH*\frac{1gH}{1molH} =0.1900gO$

4. Compute the moles of oxygen by using its molar mass:

$n_O=0.1900gO*\frac{1molO}{16gO}=0.01188molO$

5. Divide the moles of carbon, hydrogen and oxygen by the moles of oxygen (smallest one) to find the subscripts in the empirical formula:

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6. Search for the closest whole number (in this case multiply by 2):

$C_3H_6O_2$

Moreover, the empirical formula suggests this compound could be carboxylic acid since it has two oxygen atoms, nevertheless, this is not true since the molar mass is 222.27 g/mol, therefore, we should compute the molar mass of the empirical formula, that is:

$M=12*3+1*6+16*2=74g/mol$

Which is about three times in the molecular formula, for that reason, the actual formula is:

$C_9H_{18}O_6$

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